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In: Chemistry

What is the pH of a solution prepared by mixing 60.0 mL of 0.20M potassium carbonate...

What is the pH of a solution prepared by mixing 60.0 mL of 0.20M potassium carbonate (K2CO3) with 140.0 mL of 0.1M potassium bicarbonate (KHCO3)? a. 6.43 b. 6.30 c. 6.67 d. 10.32 e. 10.18

Solutions

Expert Solution

Answer – Given, [K2CO3] = [CO32-] = 0.20 M

[KHCO3] = [HCO3-] = 0.10 M

Moles of CO32- = 0.20 M * 0.060 L

                         = 0.012 moles

Moles of HCO3- = 0.10 M * 0.140 L

                            = 0.014 moles

New concentration

Total volume = 140+60 = 200 mL

[CO32-] = 0.012 moles / 0.200 L = 0.06 M

[HCO3-] = 0.0140 / 0.200 L = 0.07 M

We know the pKa2 for H2CO3 = 10.25

We know the Henderson Hasselbalch equation –

pH = pKa2 + log [CO32-] / [HCO3-]

     = 10.25 + log 0.06 / 0.07

     = 10.18

So the pH of a solution prepared by mixing 60.0 mL of 0.20M potassium carbonate (K2CO3) with 140.0 mL of 0.1M potassium bicarbonate (KHCO3) is e ) 10.18


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